# Stagnation pressure on an accelerating surface

The usual stagnation pressure $$P_0$$ on an object (e.g., a sphere) moving at velocity $$U$$ is simply

$$P_0 = P_\infty + \frac{\rho}{2}U^2$$

where $$P_\infty$$ is the pressure of the free stream.

If the object is accelerating, thus having a time dependent velocity $$U(t)$$, is the stagnation pressure given by the following equation?

$$P_0 = P_\infty + \frac{\rho}{2}U(t)^2$$

Or must we use the unsteady Bernoulli equation?

## 1 Answer

The expression for stagnation pressure is derived directly from Bernoulli's principle. Let's go through the derivation of Bernoulli's principle to understand the assumptions involved.

Consider a pipe whose axis is along the x-axis of the cartesian system. Let $$A$$ be its cross-sectional area and a fluid of density $$\rho$$ flow through the pipe. We will concentrate on a fluid parcel of length $$dx$$. Its volume is $$Adx$$ and hence, mass of the fluid parcel would be $$\rho Adx$$. Also, let $$dP$$ be the change in the pressure on moving a distance of $$dx$$ and $$\frac{dx}{dt}$$ be the flow velocity.

Ignoring gravity, for the time being, the only force acting on the fluid parcel is due to the pressure difference $$dP$$, $$F = -AdP$$ (negative sign occurs since the fluid moves towards regions of lower pressure). Applying Newton's second law,

$$\rho Adx \frac{dv}{dt} = -AdP$$ $$\rho \frac{dv}{dt} = -\frac{dP}{dx}$$

Here comes another assumption: the flow velocity is steady, i.e., $$v \equiv v(x)$$ and the time dependence is not explicit; but comes in because the fluid parcel is moving and the velocity is a function of position, $$v \equiv v(x(t))$$. Therefore,

$$\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt}$$ $$= \frac{dv}{dx}v = \frac{d}{dx}(\frac{1}{2}v^2 )$$

Now assuming fluid to be incompressible (constant $$\rho$$),

$$\frac{d}{dx}(\rho \frac{v^2}{2} + p) = 0$$

Integrating with respect to x,

$$\rho \frac{v^2}{2} + p = constant$$

If the fluid were to accelerate then the velocity field would be an explicit function of time, $$v \equiv v(x(t),t)$$ and $$\frac{dv}{dt} = v\frac{dv}{dx} + \frac{\partial v}{\partial t}$$ and we will not be able to proceed with the same derivation.

So, you probably should use the unsteady Bernoulli equation.